d
$\left|\begin{array}{ccc}1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{array}\right|=0$
$1(10)-1(7)+k(-1)-0$
$k=3$
For $k =3,2^{\text {nd }}$ system is
$4 x+5 y=7.......(1)$
and $7 x+8 y=10........(2)$
Clearly, they have a unique solution
$(2) -(1) \Rightarrow 3 x+3 y=3$
$\Rightarrow x+y=1$